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If $\frac{dy}{dx}=\frac{x+y+4}{x-y-6}$ then I set $x=z+1$ and $y=w-5$ to get $\frac{dw}{dz}=\frac{z+w}{z-w}$ from here it's been solved in the text (George Simmons Differential equations). My concern is I made a mistake using $\frac{dy}{dx}=\frac{dw}{dz}$. My reason is $\frac{dy}{dx}=\frac{dy}{dw}\frac{dw}{dz}\frac{dz}{dx}$. The answer I get doesn't have (x+y+4) or (x-y-6) anywhere unlike the solutions I find through wolfram alpha. My solution is $\tan^{-1}(\frac{y+5}{x-1})=\log(\sqrt{(x-1)^2+(y+5)^2}+c$.

Mars
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    It would help if you'd include what answer you do get: without that there's no solution to validate. – Semiclassical Sep 05 '21 at 02:11
  • Your answer should be $y=f(x)$, where $f$ is just some function that does not depend on $y$ – Andrei Sep 05 '21 at 02:12
  • @Semiclassical Sorry I should have included my solution. I just edited it. – Mars Sep 05 '21 at 02:14
  • @Andrei I included my solution. it isn't $y=f(x)$ but rather an implicit relationship. – Mars Sep 05 '21 at 02:17
  • Substitute $v=y/x$. Then you will get a seperable equation. – Sahan Manodya Sep 05 '21 at 03:44
  • Related: https://math.stackexchange.com/questions/1195583/how-to-proceed-for-these-two-differential-equations and https://math.stackexchange.com/questions/1692467/differential-equation and https://math.stackexchange.com/questions/2926293/a-specific-homogeneous-polar-differential-equation and probably many others. – Gerry Myerson Sep 05 '21 at 03:53
  • @GerryMyerson Thanks, the last one had a good solution, which indicated that wolfram alpha uses a different substitution. – Mars Sep 05 '21 at 05:07
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    OK. Should we close this question as a duplicate of that one? Or do you want to post a solution of your question here, based on what you learned from the other question? – Gerry Myerson Sep 05 '21 at 08:56
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    wolfram alpha uses substitution $u=\frac{x+y+4}{x-y-6}$. Then get separable ode $u'=\frac{u^3-u^2+u-1}{2x-2}$ and solution with $\arctan u$ – Aleksas Domarkas Sep 05 '21 at 11:19

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As indicated in the comments wolfram alpha uses the substitution $u=\frac{x+y+4}{x-y-6}$, which leads to a solution in terms of $u$. Which is different than the substitutions $x=z+5$ and $y=w-1$ which leads to solutions in terms of $x-5$ and $y+1$ as in my posted solution.

Mars
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